I conjectured the following for positive integers $n$
$$2\frac{(2n+1)^{2n+1}}{(2n+2)^{2n+2}} \le \frac{n^n}{(n+1)^{n+1}} \le 2\frac{(2n)^{2n}}{(2n+1)^{2n+1}}$$
It seems to be true for the (few) values I have tried and I was wondering if it is true for all $n \in \mathbb N_{>0}$? If so how might I see this?
To prove $$ 2\frac{(2n+1)^{2n+1}}{(2n+2)^{2n+2}}\le\frac{n^n}{(n+1)^{n+1}}\le2\frac{(2n)^{2n}}{(2n+1)^{2n+1}} $$ note that the left-hand inequality is equivalent to $$ \left(1+\frac1{2n+1}\right)^{2n+1}\ge\left(1+\frac1n\right)^n $$ which follows from the monotonic increase of $\left(1+\frac1n\right)^n$ proven in this answer using only Bernoulli's Inequality.
The right-hand inequality is equivalent to $$ \left(1+\frac1n\right)^{n+1}\ge\left(1+\frac1{2n}\right)^{2n+1} $$ which follows from the monotonic decrease of $\left(1+\frac1n\right)^{n+1}$ proven in the same answer, again using only Bernoulli's Inequality.
The function \begin{eqnarray*} \left( 1+ \frac{1}{x} \right)^x \end{eqnarray*} is increasing so \begin{eqnarray*} \left( 1+ \frac{1}{2n} \right)^{2n} \leq \left( 1+ \frac{1}{2n+1} \right)^{2n+1} \leq \left( 1+ \frac{1}{2n+1} \right)^{2n+2}. \end{eqnarray*} Square root this to get \begin{eqnarray*} \left( 1+ \frac{1}{2n} \right)^{n} \leq \left( 1+ \frac{1}{2n+1} \right)^{n+1}. \end{eqnarray*} This can be rearranged to give the upper bound ... \begin{eqnarray*} n \ln(1+\frac{1}{2n}) \leq (n+1) \ln (1+\frac{1}{2n+1}) \\ \end{eqnarray*} \begin{eqnarray*} n( \ln(n+\frac{1}{2}) -\ln(n)) \leq (n+1) (\ln(n+1)- \ln(n+\frac{1}{2})) \\ \end{eqnarray*} \begin{eqnarray*}(2n+1) \ln(n+\frac{1}{2}) \leq n \ln (n) +(n+1) \ln(n+1) \\ \end{eqnarray*} \begin{eqnarray*}n \ln n +(2n+1) \ln(2n+1) \leq \ln 2 +2n \ln(2n) +(n+1) \ln(n+1) \\ \end{eqnarray*} \begin{eqnarray*} \frac{n^n}{(n+1)^{n+1}} \leq 2 \frac{(2n)^{2n}}{(2n+1)^{2n+1}}. \end{eqnarray*} The lower bound can probably be shown similarly.
At least, for large values of $n$, take the logarithms and expand as series to get $$\log\left(2\frac{(2n+1)^{2n+1}}{(2n+2)^{2n+2}}\right)=-\log \left({n}\right)-1-\frac{3}{4 n}+\frac{7}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log\left(\frac{n^n}{(n+1)^{n+1}}\right)=-\log \left({n}\right)-1-\frac{1}{2 n}+\frac{1}{6 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log\left(2\frac{(2n)^{2n}}{(2n+1)^{2n+1}}\right)=-\log \left({n}\right)-1-\frac{1}{4 n}+\frac{1}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ Now, using $A=e^{\log(A)}$, we have, for large $n$, $$2\frac{(2n+1)^{2n+1}}{(2n+2)^{2n+2}}=\frac 1e \left(\frac{1}{n}-\frac{3}{4 n^2}+\frac{55}{96 n^3}\right)+O\left(\frac{1}{n^4}\right)$$ $$\frac{n^n}{(n+1)^{n+1}}=\frac 1e \left(\frac{1}{n}-\frac{1}{2 n^2}+\frac{7}{24 n^3}\right)+O\left(\frac{1}{n^4}\right)$$ $$2\frac{(2n)^{2n}}{(2n+1)^{2n+1}}=\frac 1e \left(\frac{1}{n}-\frac{1}{4 n^2}+\frac{7}{96 n^3}\right)+O\left(\frac{1}{n^4}\right)$$ Now, you can conclude since $$\frac{n^n}{(n+1)^{n+1}}-2\frac{(2n+1)^{2n+1}}{(2n+2)^{2n+2}}=\frac{1}{4 e n^2}-\frac{9}{32 e n^3}+O\left(\frac{1}{n^4}\right)$$ $$2\frac{(2n)^{2n}}{(2n+1)^{2n+1}}-\frac{n^n}{(n+1)^{n+1}}=\frac{1}{4 e n^2}-\frac{7}{32 e n^3}+O\left(\frac{1}{n^4}\right)$$
